Default `\int`

is the popular command by which the integral symbol is represented in the document. And the **bigints** package is designed to enlarge the shape of the symbol one by one.

```
\documentclass{article}
\usepackage{bigints}
\begin{document}
\[ \int f(x) dx \]
\[ \int \frac{\tan^{-1}x}{x(x^2+1)}dx \]
%Use bigints package
\[ \bigintssss f(x)dx ,\bigintsss f'(x)dx ,\bigintss f''(x)dx,\bigints \frac{f(x)}{g(x)}dx,\bigint \frac{f'(x)}{g'(x)}dx \]
\end{document}
```

**Output : **

## Integral symbol with above and below limits

You can use limits on integral symbols in two ways. First, it uses superscript and subscript with the `\int`

command, so that the value of limit will sit lightly on the right side with a symbol.

Second, to use the limits above and below the integral symbol, you need to use `\limits`

command along with `\int`

command.

```
\documentclass{article}
\begin{document}
\[ \int_5^7 \frac{x^3}{x^2-4}dx \]
\[ \int\limits_5^7 \frac{x^3}{x^2-4}dx \]
\[ \int_0^\infty \frac{\sin x}{x(1+x^2)}dx \]
\[ \int\limits_0^\infty \frac{\sin x}{x(1+x^2)}dx \]
\end{document}
```

**Output : **

Of course, you can see some differences by looking at the output above.

Instead of using so many commands, you can convert them to one command with the help of `\newcommand.`

```
\documentclass{article}
\usepackage{bigints}
\newcommand{\integral}[2]{\int\limits_{#1}^{#2}}
\begin{document}
\[ \integral{5}{7} \frac{x^3}{x^2-4}dx \]
\[ \integral{0}{\infty}\frac{\sin x}{x(1+x^2)}dx \]
\[ \integral{X_1}{X_2}a_i x^i dx= a_i \frac{X^{i+1}_2 - X^{i+1}_1}{i+1}\]
\end{document}
```

**Output : **

## Double integral symbol [Surface integral]

When you use limits with double integral, need to use the value of limit twice with `\int`

command.

```
\documentclass{article}
\usepackage{amsmath}
\newcommand{\integral}[2]{\int\limits^{#1}_{#2}}
\begin{document}
\[ \iint_S \textbf{F}.d\textbf{S}= \iint\limits_S \textbf{F.n}dS \]
\[ \integral{2\pi}{0}\integral{1}{0} \;r3\; \sin 2\theta\; dr \;d\theta \]
\[ \integral{2\pi}{v-0}\integral{1}{n-0}\langle u\sin v,1-u^2,-u\cos v\rangle.\langle 2u^2\cos v,2u^2 \sin v,u\rangle dudv \]
\end{document}
```

**Output : **

If you think, you will use a down limit then using `\iint`

for double integral is enough.

## Triple integral symbol

In this case, you need to represent it in the same way as shown above. However, the triple integral requires the use of the `\iiint`

command.

```
\documentclass{article}
\usepackage{amsmath}
\newcommand{\integral}[2]{\int\limits^{#1}_{#2}}
\begin{document}
\[ \iiint_V \nabla \cdot\textbf{F}dV \]
\[ V =\iiint_S dx\; dy\; dz\; \]
\[ \integral{2}{0}\integral{4}{x^2}\integral{2-x}{0}\; f(x,y,z)\;dz\;dy\;dx \]
\end{document}
```

**Output : **

## Closed integral symbol

Pxfonts package has been implemented for Closed Integral which contains all types of commands. The same symbol is also present in other packages like **stix**, **mathabx**, etc.

```
\documentclass{article}
\usepackage{pxfonts}
\begin{document}
\[ \oint,\ointclockwise,\ointctrclockwise \]
\[ \oiint,\oiintclockwise,\oiintctrclockwise \]
\[ \oiiint,\oiiintclockwise,\oiiintctrclockwise \]
\[ \sqint,\sqiint,\sqiiint \]
\end{document}
```

**Output : **

## Integral evaluation bar

Integration evolution bar can be defined manually. Again, you can solve the `\eval`

command from the **physics** package. Following code solves the problem in two ways.

```
\documentclass{article}
\usepackage{physics}
\begin{document}
\[\frac{x^2}{2} \bigg|_{a_1}^{a_2} \]
%Use physics package
\[ \eval{x}_0^\infty \]
\end{document}
```

**Output : **

## Square brackets with a limit

In this case, the square bracket must be larger than the expression. For this manually the `\bigg`

command has been used. With this, you can pass the value in superscript and subscript to give a limit.

```
\documentclass{article}
\usepackage{physics}
\begin{document}
\[ \bigg[\frac{x^2}{2} \bigg]_{a_1}^{a_2} \]
\[ \bqty\bigg{x}_{-\infty}^{+\infty} \]
\end{document}
```

**Output : **